Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  long long a[n][m], c[n][m];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) cin >> a[i][j];
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      c[i][j] = a[i][j];
    }
  }
  vector<long long> row, col, row1, col1;
  for (long long i = 0; i < n; i++) {
    long long mn = 501;
    for (long long j = 0; j < m; j++) {
      if (a[i][j] < mn) mn = a[i][j];
    }
    for (long long k = 0; k < mn; k++) {
      row.push_back(i + 1);
    }
    for (long long j = 0; j < m; j++) {
      a[i][j] = a[i][j] - mn;
    }
  }
  for (long long i = 0; i < m; i++) {
    long long mn = 501;
    for (long long j = 0; j < n; j++) {
      if (a[j][i] < mn) mn = a[j][i];
    }
    for (long long k = 0; k < mn; k++) col.push_back(i + 1);
    for (long long j = 0; j < n; j++) {
      a[j][i] = a[j][i] - mn;
    }
  }
  long long flag = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      if (a[i][j] != 0) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  if (flag == 1)
    cout << "-1";
  else {
    for (long long i = 0; i < m; i++) {
      long long mn = 501;
      for (long long j = 0; j < n; j++) {
        if (c[j][i] < mn) mn = c[j][i];
      }
      for (long long k = 0; k < mn; k++) col1.push_back(i + 1);
      for (long long j = 0; j < n; j++) {
        c[j][i] = c[j][i] - mn;
      }
    }
    for (long long i = 0; i < n; i++) {
      long long mn = 501;
      for (long long j = 0; j < m; j++) {
        if (c[i][j] < mn) mn = c[i][j];
      }
      for (long long k = 0; k < mn; k++) {
        row1.push_back(i + 1);
      }
      for (long long j = 0; j < m; j++) {
        c[i][j] = c[i][j] - mn;
      }
    }
    if (row.size() + col.size() <= row1.size() + col1.size()) {
      cout << row.size() + col.size() << "\n";
      if (row.size() + col.size() == 0) return;
      for (long long i = 0; i < row.size(); i++)
        cout << "row"
             << " " << row[i] << "\n";
      for (long long i = 0; i < col.size(); i++)
        cout << "col"
             << " " << col[i] << "\n";
    } else {
      cout << row1.size() + col1.size() << "\n";
      if (row1.size() + col1.size() == 0) return;
      for (long long i = 0; i < row1.size(); i++)
        cout << "row"
             << " " << row1[i] << "\n";
      for (long long i = 0; i < col1.size(); i++)
        cout << "col"
             << " " << col1[i] << "\n";
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}